This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.
32300 = a. 750 or a = 32300/750 = 43 square inches.
As the beam is 16" deep, this would mean an additional thickness
- 43/16 = 2 11/16
Adding this to the 7 1/4" already found to be necessary, we have
7 1/4 + 2 11/16 = 6 15/16 or the beam would need to be, say 10" x 16".
Now the size of B C must be made sufficient not to crush in the soft underside of the beam at B. The bearing here would be across the fibres of the beam, and we find (Table IV) that the safe compressive stress of Georgia pine across the fibres is
(c/f) = 200 pounds. We need therefore an area a = 25500/200 =128 inches.
As the beam is only 10" wide the strut B C will have to measure,
128/10 = 12 4/5 inches the other way, or we will say it could be 10" x 12".
This strut itself might be made of softer wood than Georgia pine, say of spruce; the average compression on it is
25500/10.12 =212 pounds per square inch.
Now spruce will stand a compression on end (Table IV) of (c/f) = 650 or, even if spruce is used, the actual strain would be less than one-third of the safe stress. At the foot of the strut B C we put an iron plate, to prevent the rod from crushing in the wood. The rod itself must bear on the plate at least
25500/12000 = 2,1 square inches, or it would crush the iron - (12000 pounds being the safe resistance of wrought-iron to crushing).
The safe tensional stress of wrought-iron being
12000 pounds per square inch (Table IV), we have the necessary area for tie-rod A C from Formula (6)
34638 = a. 12000 or a= 34628/12000= 2,886 square inches.
From a table of areas we find that we should require a rod of 1 15/16" diameter, or say a 2" rod.
The area of a 2" rod being =3,14 square inches the actual tensional stress, per square inch on the rod, will be only
34638/3,11= 11312 pounds per square inch.
Size of Strut.
Iron Shoe to
Size of Tie-rod.
We must now proportion the bearing of the washer at "A" end of tie-rod. The amount of the crushing coming on washer will be whichever of the two strains at A, (viz. B A and A C) is the lesser, or D A in our case, which is 32300 pounds. We must therefore have area enough to the washer not to crush the end of beam (or along its fibres), the safe resistance of which we already found to be: (c/f) = 750 pounds per square inch; we need therefore
32300/750= 43 square inches.
The washer therefore should be about 6 1/2 by 6 1/2"
Size of Washer.
The end of the rod must have an "upset" screw-end; that is, the threads are raised above the end of rod all around, so that the area at the bottom of sinkage, between two adjoining threads, is still equal to the full area of rod. If the end is not "upset" the whole rod will have to be made enough larger to allow for the cutting of the screw at the end, which would be a wilful extravagance.
It is unnecessary to calculate the size of nuts, heads, threads, etc., as, if these are made the regulation sizes, they are more than amply strong. It should be remarked here that in all trussed beams, if there is not a central swivel, for tightening the rod, that there should be a nut at each end of the rod; and not a head at one end and a nut at the other. Otherwise in tightening the rod from one side only it is apt to tip the strut or crush it into the beam on side being tightened. We must still however calculate the verti-cal shearing across the beam at the supports, which we know equals the reaction, or 20400 pounds at each end. To resist this we have 10" x 16" =160 square inches, less 3" x 16", cut out to allow rod end to pass, or say 112 square inches net, of Georgia pine, across the grain; and as (g/f) = 570 pounds per square inch (see Table IV); the safe vertical shearing stress at each support would be (Formula 7) 112.570 = 63840 pounds or more than three times the actual strain. Then, too, we should see that the bearing of beam is not crushed. It bears on each reaction 16 inches, or has a bearing area = 16.10 = 160 square inches.
(c/f) for Georgia pine, across the fibres, Table IV, is
(c/f) = 200, therefore the beam will bear safely at each end
160.200 = 32000 pounds or about one-half more than the reaction. There will be no horizontal shearing, of course, except in that part of beam under transverse strain, and this certainly cannot amount to much. The beam is therefore amply safe.
Now let us calculate the deflection. The modulus of elasticity for Georgia pine, Table IV is: e = 1200000 pounds-inch. The average compression strain in A F was 750 pounds per square inch, therefore the amount of contraction (Formula 88)1 x = 750.304/12000000 = 0,19 inches.
Now A D (in Fig. 146) will be one-half of this, or 0,095 inches.
The amount of elongation in A C will be, remembering that we found the average stress to be only 11312 pounds per square inch, and that for wrought-iron e = 27000000 (Formula 88) x = 11312.163/27000000 = 0,0682
The exact length of A C (Fig. 147 should be 163,41 not 163"). Therefore D F (Fig. 146) will be
D F=163,41 + 0,0682 = 163,4782" DB = 152 - 0,19 = 151", 81
Therefore (Fig. 146)
Deflection of Beam.
= 60", 655
Now B C (Fig. 147) would be = 60 , deducting this from the above we should have a deflection = 0",655.
To this we must add the contraction of B C. The strut will be less than 60" long, say about 50". The average compressive stress per square inch we found = 212 pounds. The modulus of elasticity
1 In reality the contraction of A F would be much less, as the part figured for transverse strain only would very materially help to resist the compression, one-half of it being in tension.
for spruce, Table IV, is e = 850000, therefore contraction in strut (Formula 88) x = 212.50/850000 = 0,0125